One of the things I get asked about quite often is "what the heck is
crh and how is it calculated." Nearly every time I have been asked
that question, I've wound up expanding the definition in my "Definitions
and Information" so that it is now one of the longest in those listings.
However, even this explanation does not seem to be detailed enough, as
I still get questions about it. So, here is a longer, more complete
explanation.

Let's start with my Definition:

crh
- Caliber Radius Head. The pointed head of a projectile is described
in terms of its ballistic length and the radius of the curvature of its
nose. Larger numbers mean a more streamlined profile. Properly,
crh is shown as a dual number such as 3/4crh, with the first number indicating
the ballistic length and the second number indicating the radius of the
curvature, but it is often abbreviated to a single number such as 4crh.
In the sketch at the right, the dotted line between Points A and B is the
"shoulder" which is the start point of the nose and the distance between
these points is the caliber of the projectile. In this sketch, the
radius of the curvature is from Point A to Point E and is four times the
caliber of the projectile. The vertical distance between Points C
and D is the ballistic length and is the most important factor in the design
of a shell for stability in flight. In this sketch, the ballistic
length is 4, as Point E is on the same plane as Points A and B. From
these numbers, this projectile would properly be described as 4/4crh
but this would normally be abbreviated to just 4crh. Shells
of this general shape are described as being "ogival headed" and have superior
ballistic performance. As can easily be imagined, a 6crh shell is
more pointed and streamlined than is a 4crh shell. When crh is described
as 5/10crh it means that the radius is 10 calibers long but the
ballistic length is that of a 5crh shell. When a projectile is described
as 5/crh
it means that it has a ballistic length of 5 and its nose shape is conical
(infinite radius), not ogival. Most USN projectiles had secant ogive
ballistic nose shapes which were somewhat more conical than a simple tangent
ogive (smooth merging joint with cylindrical lower-body side) and gave
them a distinct "shoulder" where the nose met the cylindrical side of the
lower body). This shape has slightly reduced air friction compared
to a tangent ogive nose of the same length above the cylindrical body.

With that explanation in mind, let's examine the sketch above more
closely. As noted on the sketch, the radius of the ogive is 4crh.
This means that, as shown in the sketch, the ogive is part of a curve that
is 4X the caliber. Note that the center of the radius in my sketch
(Point E) is precisely in line with the start of the ogive (the "shoulder"
between Points A and B). As these three points are in a direct line, this
means that the ballistic length of the projectile in this sketch is 4.
That means that the projectile in my sketch is properly defined as being
4/4crh.

Now, if we move the center of the radius (Point E) downwards while keeping
the radius constant, this will decrease the ballistic length of the projectile
but keep the radius of the ogive the same.

Catch that? It's tricky.

Let's do a couple of mathematical and geometric examples working with
the projectiles used for the famous British 15"/42 Mark I naval cannon.

The older British 15" projectiles commonly known as the "4crh" shells
were really 3.05/4crh. Again, this meant that the ballistic length was
3.05 times the caliber while the ogive radius was 4 times the caliber.
Some quick calculations with these numbers show that the ballistic length
for this projectile was 45.75 inches and the ogive radius was 60 inches.
To show how these numbers are used, we need to use a little basic geometry.

First, the length of a chord through a circle is determined by the following
geometric formula:

chord length = 2 * [h * (2R - h)]^0.5

Where:

h = height of the chord (distance from the circle's circumference
to the chord)
R = Radius of the circle

For crh calculations, h (height) is 0.5 * the caliber of the projectile
and R is the caliber of the projectile times the ballistic length multiplier.
And, as the ballistic length of a projectile is obviously only half of
the chord length, the "2" in the front of this above formula is eliminated.
This means that the ballistic length of a projectile can be calculated
using the values for crh as follows:

Referring to the sketch, let's use the above formula for an imaginary 15"
projectile with 4/4crh dimensions:

ballistic length = [15^2 * (4 - 0.25)]^0.5
= 29"

Now, as noted above, the British 4crh 15" projectile was actually 3.05/4crh.
Thus, the ballistic length of this projectile would be:

ballistic length = [15^2 * (3.05 - 0.25)]^0.5
= 25.1"

So, the ballistic length of the 3.05/4crh projectile was about 4 inches
shorter than a "true" 4crh projectile of 4/4crh.

As the overall length of the British 3.05/4crh projectile was about
56", we can quickly calculate that the shoulder to base dimension of this
projectile was 56" minus 25.1" equals 30.9". This base to shoulder
dimension will be used in a subsequent calculation.

Now, let's turn our attention to the later British projectiles which
were commonly known as the "6crh" shells. Now, how they came up with
that designation is beyond me, as, unlike the "4crh" projectiles, there
is no "6" in the actual crh values. Perhaps some trivia hunter looking
though PRO files will enlighten me someday. Anyway, this projectile was
actually 5/10crh. So, it had a ballistic length of 5 and an ogive
radius of 10.

Determining the ballistic length with the formula above:

ballistic length = [15^2 * (5 - 0.25)]^0.5
= 32.7"

These numbers show that this projectile was about 7.5 inches longer than
the older 4crh projectile. Assuming that the shell body for the 6crh
projectile was roughly the same as that for the 4crh projectile, then its
total length would be 32.7 inches + 30.9 inches = 63.6 inches. This
turns out to be almost exactly the length of this projectile, thus confirming
the accuracy of these calculations.

If you are still confused about what the crh values mean at this time,
then my best suggestion is to get a compass - the drawing kind, not the
magnetic kind - and start making curves. One of my correspondents
on this matter used a mathcad program, ProE, and was able to teach himself
how these values affect projectile design. But, assuming that all
you have is a compass and a sheet of paper, let me give you an example.

Working with the dimensions for the 3.05/4crh 15" projectile, let's
start by drawing a vertical rectangle representing the shell body.
As calculated above, the shell body for this projectile was 15" x 30.9".
Unless you have a really big piece of paper and some oversized drafting
tools, this is difficult size to handle, so let's do a 1/10th scale drawing
and make our vertical rectangle 1.5" x 3.1". Similar to the sketch
above, label the corners at top end of the rectangle as Points A and B
to denote the shoulder. As the 3.05 ballistic length of this projectile
was calculated above as being 45.75 inches long, a 1/10th scale of this
value would be about 4.6 inches.

Draw the following two curves lightly, as we will be erasing them in
the next step. Using the compass, draw a curve with a radius of 4.6
inches starting at Point A with the center of the radius (Point E) precisely
in line with the shoulder. Repeat this procedure on the opposite
side of the rectangle and draw a curve starting at Point B, so that you
now have two curves intersecting above the rectangle. Label this
intersection as Point D. Your drawing should now look similar to
the sketch above and represents a ballistic length of 3.05 and an ogive
radius of 3.05.

Now, the next step is to give your projectile an ogive radius of 4.
So, erase the two curves you have just drawn, but keep your Point D.
As the ogive radius of this projectile was 4 times the caliber or 60 inches,
a 1/10th scale value means that you should now set your compass to draw
a 6 inch curve. Then, draw a curve that starts at Point A and ends
at Point D. Note that in order to draw this curve correctly, you
must move the center of the radius (Point E) downwards, below the shoulder
line.

Repeat this procedure on the opposite side of the rectangle to finish
the drawing.